MHT-CET Physics · Thermal Properties of Matter

Conduction Through Rods and Slabs

Heat flows along a rod at the rate Q/t = KAΔT/L, which is the temperature difference divided by the thermal resistance L/KA; slabs in series add their resistances and rods side by side add their conductances, exactly like electrical resistors.

All Thermal Properties of Matter notes MHT-CET Physics strategy

Why this matters

8 PYQs, none HARD: K from the flow rate, a rod's thermal resistance, a composite slab of two materials, two rods joined end to end and then side by side, and what happens when every dimension of a rod doubles. One asks which everyday heating is convection. One card.

Concept 1 of 1: Rate of Flow and Thermal Resistance

Heat flow through a rod is like current through a resistor: the temperature difference drives it, and the rod resists with R = L/KA. Two slabs in series carry the same heat, so their resistances add; two rods side by side share the temperature difference, so their conductances add. Doubling every dimension of a rod doubles the length and quadruples the area, so the flow doubles. Conduction moves heat through a solid, convection by moving fluid (a room heater warms air that circulates), and radiation needs no medium at all.

Definition

$\dfrac{Q}{t} = \dfrac{KA\,\Delta T}{L}$ , so $K = \dfrac{QL}{tA\,\Delta T}$ .
Thermal resistance $R = \dfrac{L}{KA} = \dfrac{\Delta T}{Q/t}$ (40 °C, 1600 cal/s ⇒ 0.025 °C s/cal).
Series slabs: $R = R_1 + R_2$ (K, x and 2K, 4x ⇒ $\dfrac{3x}{KA}$ , a factor of 1/3).
Two identical rods: end to end twice the resistance, side by side half — the time for the same heat falls 4 times (12 s ⇒ 3 s).
All dimensions doubled: $A \to 4A$ , $L \to 2L$ , flow doubles.

Conduction

\frac{Q}{t} = \frac{KA\,\Delta T}{L} = \frac{\Delta T}{R}, \qquad R = \frac{L}{KA}

Worked example

A copper rod (K = 400 W/m K) 0.5 m long with cross-section 2 × 10⁻⁴ m² has its ends at 100 °C and 0 °C. Rate of heat flow?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q31Moderate

Example 1 · Thermal Properties of Matter · Heat Conduction and Thermal Resistance

A composite slab consists of two materials having coefficient of thermal conductivity

K

and

2K

, thickness

x

and

4x

respectively. The temperature of the two outer surfaces of a composite slab are

T_{2}

and

T_{1}

and

T_{2}>T_{1}

. The rate of heat transfer through the slab in a steady state is

\frac{A(T_{2}-T_{1})K}{x}\cdot f

where

f

is equal to

Adding conductivities for slabs in series

Slabs in series carry the same heat and add their RESISTANCES, x/KA + 4x/2KA = 3x/KA. Averaging K is wrong.

Scaling the flow with the area alone

Doubling all dimensions multiplies A by 4 but also doubles L, so the flow only doubles.

Drill 7 more on rate of flow and thermal resistance

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Rate of Flow and Thermal Resistance
Conduction
$\frac{Q}{t} = \frac{KA\,\Delta T}{L} = \frac{\Delta T}{R}, \qquad R = \frac{L}{KA}$

Watch out for (2)

Adding conductivities for slabs in series
→ Rate of Flow and Thermal Resistance
Scaling the flow with the area alone
→ Rate of Flow and Thermal Resistance

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